# Finitely generated module over finitely generated algebra

Question

Let $$A$$ be a finitely generated $$R$$-algebra generated by $$x_1,...,x_n$$ and $$M$$ a finitely generated $$A$$-module. If $$x_1,...,x_n\in \sqrt{\operatorname{Ann}_{A}{M}}$$, show that $$M$$ is a finitely generated $$R$$-module.

Attempt

We can replace $$A$$ by $$R[x_1,x_2....x_n]/I$$ for some ideal $$I$$ in $$R[x_1,x_2....x_n]$$. Let $$M$$ be finitely generated by $$m_1,m_2.....m_k$$ over $$A$$. Then for any $$m\in M$$, there exist $$(f_j+I)_{j=1....k}$$ in $$A$$ such that $$m=\sum_{j=1}^{k}(f_j+I).m_j$$.

Now if $$x_1,x_2...,x_n\in {\operatorname{Ann}_{A}(M)}$$, then we would be done as then $$m=\sum_{j=1}^{k}(c_j.m_j)$$, where $$c_j$$ are constant terms of $$f_j$$.However I am not able to prove it when we only know that $$x_1,x_2...,x_n\in \sqrt{\operatorname{Ann}_{A}{M}}$$.

Any help is appreciated.

$$\exists \ N\in \mathbb N$$ such that $$x_i^N\in \operatorname{Ann}_AM \ \forall \ i$$.

Choose a set of generators $$\{m_\alpha\}$$ of $$M$$ over $$A$$. I claim $$M$$ is generated over $$R$$ by $$\{x^\beta m_\alpha\}$$ where $$\beta$$ is a multi-index i.e. $$\beta=(\beta_1,\dots,\beta_n)\in \mathbb Z_{\geq 0}^n$$ such that $$|\beta|=\sum_i\beta_i\leq Nn$$. Clearly this is a finite set.

Pick any $$m\in M$$. Write $$m=\sum_\alpha c_\alpha m_\alpha$$ with $$c_\alpha\in A$$

Each $$c_\alpha$$ has a representation of the form $$c_\alpha =\sum_\beta r_{\beta,\alpha}x^\beta$$ where $$r_{\beta,\alpha}\in R$$

Now observe that $$c_\alpha m_\alpha =\sum_{\beta} r_{\beta,\alpha}x^\beta m_\alpha=\sum_{\beta : |\beta|\leq Nn} r_{\beta,\alpha}x^\beta m_\alpha$$ since the other terms will be in the annihilator of $$M$$.

Thus you see any element is a finite $$R$$ linear combinations of the generators we chose.

• I didn't think that generating set would be of that type. Thanks! May 16 '20 at 11:01